積分解説1

今回はこの積分botの積分
https://twitter.com/integralsbot/status/1429257154758774785?s=21
を解説します。

$\displaystyle\int_0^\pi\log(2\cos\frac{x}{2})\log(2\sin\frac{x}{2})dx$
$=\displaystyle\log^2{2}\int_0^\pi dx+\log2\int_0^\pi(\log\cos\frac{x}{2}+\log\sin\frac{x}{2})dx+\int_0^\pi\log\cos\frac{x}{2}\log\sin\frac{x}{2}dx$
$=\displaystyle \pi\log^2{2}+\left.4\log2・\frac{1}{4}\frac{\partial}{\partial s}B\right|_{(s,t)=(\frac{1}{2},\frac{1}{2})}+\left.2・\frac{1}{8}\frac{\partial^2}{\partial s\partial t}B\right|_{(s,t)=(\frac{1}{2},\frac{1}{2})}$

$\displaystyle\frac{\partial}{\partial s}B=\frac{\Gamma(s)\Gamma(t)}{\Gamma(s+t)}(\psi(s)-\psi(s+t))$
$(s,t)=(\frac{1}{2},\frac{1}{2})$に於いて$\displaystyle-2π\log2$
また、
$\displaystyle\frac{\partial^2}{\partial s\partial t}B=\frac{\Gamma(s)\Gamma(t)}{\Gamma(s+t)}((\psi(s)-\psi(s+t))(\psi(t)-\psi(s+t))-\psi^{(1)}(s+t))$
これは$(s,t)=(\frac{1}{2},\frac{1}{2})$で
$\displaystyle4\pi\log^2{2}-\frac{\pi^3}{6}$となる
よって、求める積分の値は$\displaystyle-\frac{\pi^3}{24}$